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Math Help - Measure Theory -- Integration/Continuity

  1. #1
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    Measure Theory -- Integration/Continuity

    Hey, I am kind of stumped on this one. Anyone want to offer hints?

    If f is in L1(m) and F(x) = integral ( f(t)dt ) from -infinity to x, then F is continuous on R.

    I know a couple things, sure:

    **F in L1(m) means that {x : f(x) = infinity} is a null set.
    **F is integrable (and therefore continuous) is integral(F(x)) < infinity.
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  2. #2
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    Quote Originally Posted by joeyjoejoe View Post
    Hey, I am kind of stumped on this one. Anyone want to offer hints?

    If f is in L1(m) and F(x) = integral ( f(t)dt ) from -infinity to x, then F is continuous on R.

    I know a couple things, sure:

    **F in L1(m) means that {x : f(x) = infinity} is a null set.
    **F is integrable (and therefore continuous) is integral(F(x)) < infinity.
    You should read again your definitions: f\in L^1(m) means that f is measurable and integrable with respect to the measure m: \int |f(x)| m(dx)<\infty. And integrable is not linked with continuous.

    To show that the function F:x\mapsto\int_{-\infty}^x f(t)dt is continuous at x\in\mathbb{R}, notice that, if y>x, F(y)-F(x)=\int_x^y f(t)dt = \int_{\mathbb{R}} {\rm 1}_{[x,y]}(t)f(t)dt (where the 1 is the indicator function). Then you can use the dominated convergence theorem, when y tends to x, to show that the previous integral converges to 0.
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